= Solution
Write $Q=(Z\times I)/\mathcal R$. Since $Q$ is <homeomorphic> to the <circle>, it is <Hausdorff>. For fixed $t\in I$, the <function>
$$
Z\longrightarrow Q,\qquad x\longmapsto p(x,t)
$$
is <continuous>, being a coordinate slice of the <quotient map>. Part (a)(i) therefore makes it constant. Choose $x_0\in Z$ and set
$$
g:I\longrightarrow Q,\qquad g(t)=p(x_0,t).
$$
Then $g$ is <continuous> and
$$
p(x,t)=g(t)\qquad\text{for every }x\in Z,\ t\in I.
$$
Since the <quotient map> $p$ is a <surjection>, so is $g$. This is <Hausdorff factorization through a cofinite product factor>: a <continuous map> into a <Hausdorff space> cannot distinguish the points of the coordinate with the <cofinite topology>.
If $g$ were injective, it would be a <continuous bijection> from the <compact> <closed interval> $I$ to the <Hausdorff space> $Q$. Part (a)(ii) would make it a <homeomorphism>, contradicting the fact that the <circle and interval are not homeomorphic>.
To recall that obstruction explicitly, remove the interior point $1/2$ from $I$. The resulting <topological subspace> is separated by the nonempty relatively <open sets> $[0,1/2)$ and $(1/2,1]$, so it is not <connected>. The <circle> with any one point removed is <connected>: after a rotation, it is the <image> under a <continuous map> of the <connected> <open interval> $(0,2\pi)$ under $t\mapsto(\cos t,\sin t)$. A <homeomorphism> and its restriction preserve <connectedness>, giving a contradiction.
Thus $g$ is not injective. Choose distinct $a,b\in I$ with $g(a)=g(b)$. Then
$$
\boxed{p(x,a)=g(a)=g(b)=p(x,b)\quad\text{for every }x\in Z.}
$$
In particular the requested $x,a,b$ exist. The argument proves the stronger conclusion that the same two interval coordinates work for every $x$.
Back to article page
